AP Calculus BC

Integral of ln 2x: Answer, Proof, and Steps

Q: Why is there no 1/2 out front the way there is for sin 2x?

Because \frac{d}{dx}[\ln 2x] = \frac{1}{x}, not \frac{2}{x}. The inside coefficient cancels inside the chain rule itself, so the antiderivative has nothing to compensate for.

Q: Can I rewrite the log before integrating?

Yes, and it is a clean shortcut. On the domain x > 0, \ln 2x = \ln 2 + \ln x, so the integral is x\ln 2 + x\ln x - x + C, which factors back to x\ln 2x - x + C.

Q: What is the definite integral from 1 to 2?

Evaluate x\ln 2x - x at both ends: (2\ln 4 - 2) - (\ln 2 - 1) = 3\ln 2 - 1, roughly 1.079.

The integral of ln(2x) is x ln(2x) - x + C, valid for x > 0. Integration by parts with u = ln(2x) and dv = dx works because the derivative of ln(2x) is 1/x, which cancels against the x from v and leaves the integral of 1. No factor of 1/2 appears.

\int \ln 2x\,dx = x\ln 2x-x + C

Parts with nothing to integrate

There is no antiderivative of $\ln 2x$ to quote and no second factor to pair it with. The move is to supply one: let $dv = dx$ , so $v = x$ , and let $u$ be the entire integrand.

u = \ln 2x, \quad du = \frac{1}{x}\,dx, \quad dv = dx, \quad v = x

That $du$ deserves a second look. The chain rule gives $\frac{d}{dx}[\ln 2x] = \frac{2}{2x} = \frac{1}{x}$ , so the inside 2 cancels against itself and $\ln 2x$ has the same derivative as $\ln x$ .

Running the formula

In $uv - \int v\,du$ the $x$ from $v$ meets the $\frac{1}{x}$ from $du$ , and the new integrand is just 1.

\int \ln 2x\,dx = x\ln 2x - \int x\cdot\frac{1}{x}\,dx = x\ln 2x - \int 1\,dx

\int \ln 2x\,dx = x\ln 2x - x + C

Check by differentiating: the product rule gives $\ln 2x + x\cdot\frac{1}{x} - 1 = \ln 2x + 1 - 1 = \ln 2x$ .

How it relates to the integral of ln x

For $x > 0$ , $\ln 2x = \ln 2 + \ln x$ , so this answer sits a distance $x\ln 2$ away from $x\ln x - x$ . That gap is a function of $x$ , not a constant, which is why the two integrals genuinely differ even though the integrands differ by a constant.

The mistake students make

The frequent error is dividing by 2, writing $\frac{x\ln 2x - x}{2}$ by analogy with $\int \sin 2x\,dx = -\frac{\cos 2x}{2}$ . That division applies when the inside coefficient survives differentiation and needs cancelling. In a logarithm it cancels on its own, since $\frac{d}{dx}[\ln 2x] = \frac{1}{x}$ , so there is nothing left to undo.

The second slip is dropping the $-x$ and answering $x\ln 2x$ . Differentiating that gives $\ln 2x + 1$ , overshooting by exactly the 1 that the $-x$ term exists to cancel.

The third is ignoring the domain. $\ln 2x$ exists only for $x > 0$ , so $x\ln 2x - x$ is an antiderivative on $(0, \infty)$ and nowhere else. A definite integral reaching down to 0 is improper (Topic 6.13) and has to be evaluated as a limit rather than by plugging the endpoint in. Absolute value bars do not extend it either, the way they do for $\int\frac{dx}{x}$ : the integrand itself has no negative branch to repair.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

Why is there no 1/2 out front the way there is for sin 2x?

Because $\frac{d}{dx}[\ln 2x] = \frac{1}{x}$ , not $\frac{2}{x}$ . The inside coefficient cancels inside the chain rule itself, so the antiderivative has nothing to compensate for.

Can I rewrite the log before integrating?

Yes, and it is a clean shortcut. On the domain $x > 0$ , $\ln 2x = \ln 2 + \ln x$ , so the integral is $x\ln 2 + x\ln x - x + C$ , which factors back to $x\ln 2x - x + C$ .

What is the definite integral from 1 to 2?

Evaluate $x\ln 2x - x$ at both ends: $(2\ln 4 - 2) - (\ln 2 - 1) = 3\ln 2 - 1$ , roughly $1.079$ .